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Approaching the following problem:

A plane monochromatic electromagnetic wave with wavelength $\lambda = 2.0 cm$, propagates through a vacuum. Its magnetic field is described by $ > \vec{B} = ( B_x \hat{i} + B_y \hat{j} ) \cos(kz + ωt) $, where $B_x = 1.9 \times 10^{-6} T, B_y = 4.7 \times 10^{-6} T$, and $\hat i$ and $\hat j$ are the unit vectors in the $+x$ and $+y$ directions, respectively. What is $S_z$, the $z$-component of the Poynting vector at $(x = 0, y = 0, z = 0)$ at $t = 0$?

Question 1

What is the Poynting vector in a general sense? i.e. Abstractly, what am I trying to compute?

I know it is described as the directional energy flux density of an electromagnetic field. But, it is not traveling through a symmetric surface or anything of this nature so what am I trying to compute here?

Question 2

How do I actually compute the value I am looking for?

I know $\vec S \equiv \frac{\vec E \times \vec B}{ \mu_0}$ where $\vec E$ is the electric field, $\vec B$ is the magnetic field, and $\mu_0 = 4 \pi \times 10^{-7}$. But, I was under the impression that the magnetic field and the electric field are always perpendicular, why isn't $S_z = \frac{ c}{ \mu_0}$ since we are in a vacuum and $E = cB$?

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1 Answer 1

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The Poynting vector $\mathbf{S}$ represents the flow of energy in an EM field. Specifically, if $u$ is the energy density of the field, the Poynting vector satisfies the continuity equation for it: $$\frac{\partial u}{\partial t}+\nabla\cdot\mathbf{S}=0$$ in vacuum. (This is Poynting's theorem.)

In your particular problem, $E$ and $B$ are perpendicular and their cross product is proportional to the product of their amplitudes. Thus $$S_z={c\over\mu_0}B^2.$$ You then have to use your knowledge of $B$ to work out $S$.

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  • $\begingroup$ How do I approach computing $E$, at $x = 0, y = 0, z = 0, t = 0$, wouldn't $B = B_x + B_y$, hence $E = c(B_x + B_y)$? (using this value doesn't give me the correct result) $\endgroup$
    – rudolph9
    Jul 28, 2012 at 4:34
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    $\begingroup$ Yes, $E$ you work out from $E=cB$. Thus $S_z={c\over\mu_0}(B_x^2+B_y^2)$. $\endgroup$
    – Emilio Pisanty
    Jul 28, 2012 at 9:44
  • $\begingroup$ It ended up being $S_z = -\frac{ c}{ \mu_0}(B_x^2 + B_y^2)$, a negative, and I do not understand why? How does the cross product work out... What I am I interpreting incorrect in the following: $\vec E = [c B_x, cB_y, 0], \vec B = [B_x, B_y, 0]$ and $\vec E \times \vec B = i(c B_y \cdot 0 - 0 \cdot B_y) - j(c B_x \cdot 0 - 0 \cdot B_x) + k(c B_x \cdot B_y - B_x \cdot B_y) = 0$ ? $\endgroup$
    – rudolph9
    Jul 28, 2012 at 14:09
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    $\begingroup$ $E$ and $B$ are proportional in magnitude but orthogonal to each other as well as to the wave vector, and such that $E$, $B$ and $k$ are a right-handed system. Since $E$, $B$ and $E\times B $ are also right-handed, $S$ goes in the direction of $\vec{k}$ which in this case is negative $z$. $\endgroup$
    – Emilio Pisanty
    Jul 28, 2012 at 16:38

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